An object of mass `m` is allowed to fall from rest along a rough inclined plane. The speed of the object on reaching the bottom of the plane is proportional to:

Options

A. `m^(0)`
B. m
C. `m^(2)`
D. `m^(-1)`

Answer

To determine the speed of an object of mass `m` falling from rest along a rough inclined plane, we can use the principles of energy and the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on the object
When the object is on the inclined plane, the forces acting on it are:
- The gravitational force, \( mg \), acting downwards.
- The normal force, \( N \), acting perpendicular to the surface of the incline.
- The frictional force, \( F_f \), acting opposite to the direction of motion.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( N = mg \cos(\theta) \)
- Parallel to the incline: \( F_{\text{parallel}} = mg \sin(\theta) \)

### Step 3: Determine the frictional force
The frictional force can be expressed as:
\[ F_f = \mu N = \mu mg \cos(\theta) \]
where \( \mu \) is the coefficient of friction.

### Step 4: Apply the work-energy theorem
According to the work-energy theorem:
\[ \text{Work done by net force} = \text{Change in kinetic energy} \]
The work done by the gravitational force and the frictional force can be expressed as:
\[ \text{Work done by gravity} - \text{Work done by friction} = \Delta KE \]

### Step 5: Calculate the work done
The height of the incline is \( h \), and the length of the incline can be expressed in terms of height and angle:
\[ r = \frac{h}{\sin(\theta)} \]
The work done by gravity is:
\[ W_g = mg h \]
The work done against friction is:
\[ W_f = F_f \cdot r = \mu mg \cos(\theta) \cdot \frac{h}{\sin(\theta)} = \mu mg h \cot(\theta) \]

### Step 6: Set up the equation
Now, we can set up the equation:
\[ mg h - \mu mg h \cot(\theta) = \frac{1}{2} mv^2 \]

### Step 7: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g h - \mu g h \cot(\theta) = \frac{1}{2} v^2 \]

### Step 8: Solve for \( v^2 \)
Rearranging gives:
\[ v^2 = 2gh(1 - \mu \cot(\theta)) \]

### Step 9: Find the speed \( v \)
Taking the square root gives:
\[ v = \sqrt{2gh(1 - \mu \cot(\theta))} \]

### Conclusion
The speed of the object on reaching the bottom of the inclined plane is proportional to:
\[ \sqrt{h} \]
since the other factors are constants or depend on the angle and friction.